NCERT Class 11 Physics Example 5.7 Solution | Vertical Circular Motion

 - Dr.Sanjaykumar pawar

NCERT Class 11 Physics Example 5.7 showing a bob performing vertical circular motion with speeds at points A, B, and C.

Internal Links

• NCERT Class 11 Physics Chapter 5 Work, Energy and Power Notes
• Work-Energy Theorem Explained
• Conservation of Mechanical Energy
• Circular Motion Complete Notes
• Centripetal Force Formula and Examples
• Vertical Circular Motion Theory
• NCERT Class 11 Physics All Solved Examples
• NCERT Chapter 5 Exercise Solutions
• Important JEE Questions on Circular Motion
• NEET Physics Circular Motion Practice Questions
• Class 11 Physics Formula Sheet
• Motion in a Vertical Circle Numerical Problems
• Projectile Motion Notes
• Newton's Laws of Motion Revision 
• Class 11 Physics Previous Year Questions

NCERT Class 11 Physics

Chapter 5: Work, Energy and Power

Example 5.7

Question

A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v₀ at the lowest point A such that it completes a semi-circular trajectory in the vertical plane, with the string becoming slack only on reaching the topmost point C.

Obtain expressions for:

1. The speeds at points B and C.
2. The ratio of the kinetic energies K_B/K_C.
3. Comment on the nature of the trajectory of the bob after it reaches point C.

Solution

Step 1: Energy at Point A

Take the potential energy at the lowest point A as zero.

Total Mechanical Energy at A

E = ½mv₀²

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Step 2: Forces at Point A

Applying Newton's Second Law:

T_A − mg = mv₀²/L

where T_A is the tension in the string.

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Step 3: At the Highest Point C

Since the string just becomes slack,

T_C = 0

Only gravity provides the centripetal force.

Therefore,

mg = mv_C²/L

Hence,

v_C² = gL

Speed at C:

v_C = √(gL)

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Step 4: Energy at Point C

Height of C above A = 2L

Potential Energy = 2mgL

Kinetic Energy = ½mv_C²

Substituting v_C² = gL,

KE = ½mgL

Total Energy at C:

E = ½mgL + 2mgL

E = 5/2 mgL

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Step 5: Find Initial Speed

By conservation of energy,

½mv₀² = 5/2 mgL

Therefore,

v₀² = 5gL

Initial Speed:

v₀ = √(5gL)

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Step 6: Speed at Point B

Height of B above A = L

Energy at B:

½mv_B² + mgL = ½mv₀²

Substitute v₀² = 5gL

½mv_B² = 3/2 mgL

v_B² = 3gL

Speed at B:

v_B = √(3gL)

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Step 7: Ratio of Kinetic Energies

Kinetic Energy at B:

K_B = ½m(3gL) = 3/2 mgL

Kinetic Energy at C:

K_C = ½m(gL) = ½mgL

Therefore,

K_B / K_C = 3

Ratio:

K_B : K_C = 3 : 1

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Step 8: Motion After Point C

At point C, the string becomes slack because the tension becomes zero. The velocity of the bob is horizontal. Therefore, after leaving the circular path, the bob follows the path of a projectile under the action of gravity.

Final Answers

1. Speed at B: √(3gL)
2. Speed at C: √(gL)
3. Minimum initial speed: √(5gL)
4. Ratio of kinetic energies: 3 : 1
5. After reaching C, the bob executes projectile motion.

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Practice Questions

1. What is the minimum speed required at the lowest point to complete a vertical circle?
   
   Answer: √(5gL)


2. What is the speed of the bob at the highest point?
   
   Answer: √(gL)


3. What is the speed at the side point B?
   
   Answer: √(3gL)


4. Find the ratio of kinetic energies at B and C.
   
   Answer: 3 : 1


5. Why does the string become slack at the highest point?
   
   Answer: Because the tension becomes zero and gravity alone provides the centripetal force.